Exercises — Magnetic force on current-carrying conductor

KYA: perpendicular ka matlab $\theta =90^{\circ }$ hai, isliye $\sin \theta =1$ aur formula $F=BIL$ ban jaata hai. KYUN: $90^{\circ }$ par poori wire field ko "cut across" karti hai, isliye har carrier ko maximum sideways shove milta hai. $F=BIL=0.20\times 5.0\times 0.40=0.40\mathrm{N}.$ Answer: $0.40\mathrm{N}$.

Vector form: $\vec{F}=I\vec{L}\times \vec{B}$ magnitude ke saath $F=BIL\sin \theta$.

• Maximum $\theta =90^{\circ }$ par (wire $\perp$ field): $F=BIL$.
• Zero $\theta =0^{\circ }$ ya $180^{\circ }$ par (wire field ke $\parallel$ ya anti-$\parallel$): $F=0$. KYUN zero jab parallel: $\vec{L}\parallel \vec{B}\Rightarrow \vec{L}\times \vec{B}=\vec{0}$, kyunki cross product measure karta hai ki do vectors kitne perpendicular hain.

KYA: wire tilted hai, isliye poora $\sin \theta$ rakho. KYUN: wire ka sirf woh hissa push hota hai jo field ke across hai; $\sin 30^{\circ }=0.5$ exactly bolta hai ki aadha "sideways-ness" bachta hai. $F=BIL\sin \theta =0.50\times 3.0\times 0.60\times \sin 30^{\circ }=0.90\times 0.5=0.45\mathrm{N}.$ Answer: $0.45\mathrm{N}$.

KYA: hume $F$, $I$, $L$ pata hai, aur $\theta =90^{\circ }$; $F=BIL$ ko $B$ ke liye solve karo. KYUN rearrange karein: formula chaar numbers ko baandhe rakhta hai; koi bhi teen jaanne se chauthaa mil jaata hai. $B=\frac{F}{IL}=\frac{0.24}{2.0\times 0.30}=\frac{0.24}{0.60}=0.40\mathrm{T}.$ Answer: $B=0.40\mathrm{T}$.

KYA: axes set up karo aur cross product se haath ko double-check karo. Maano $\hat{x}=$ east, $\hat{y}=$ north, $\hat{z}=$ up. Current north hai: $\vec{L}\propto \hat{y}$. Field down hai: $\vec{B}\propto -\hat{z}$. $\vec{L}\times \vec{B}\propto \hat{y}\times (-\hat{z})=-(\hat{y}\times \hat{z})=-\hat{x}=\text{west}.$ Answer: force west ki taraf point karta hai. Hand check (Fleming's Left-Hand Rule, is page ke upar define kiya): Pehli ungli neeche (field), doosri ungli north (current) → thumb west ki taraf point karti hai. ✓

KYA: $\vec{F}=I\vec{L}\times \vec{B}$ component-by-component compute karo. KYUN cross product: $\vec{B}$ ka $\hat{x}$-part wire ke parallel hai aur kuch contribute nahi karna chahiye; cross product automatically use cheezon ko khatam kar deta hai.

KYA: parent note ka uniform-field shortcut use karo. Uniform $\vec{B}$ mein, $\vec{F}=I(\int d\vec{\ell })\times \vec{B},$ aur $\int d\vec{\ell }$ sirf start end se finish end tak straight vector hai — yani diameter, length $2R$. KYUN yeh save karta hai: curve ke around force integrate karna padta nahin; sirf endpoints matter karte hain. $L_{\text{eff}}=2R=0.40\mathrm{m}.$ Diameter $\vec{B}$ ke perpendicular hai (jo page se bahar hai), isliye $\theta =90^{\circ }$: $F=BI{L}_{\text{eff}}=0.30\times 6.0\times 0.40=0.72\mathrm{N}.$ Direction — derive karo, sirf read off mat karo: diameter ko $x$-axis ke along rakho, effective current vector start end se finish end ki taraf point karte hue, ${\vec{L}}_{\text{eff}}=2R\hat{x}$. Field page se bahar hai, $\vec{B}=B\hat{z}$. Tab $\vec{F}=I{\vec{L}}_{\text{eff}}\times \vec{B}=I(2R)B(\hat{x}\times \hat{z})=I(2R)B(-\hat{y}),$ $\hat{x}\times \hat{z}=-\hat{y}$ use karke. Isliye force $-\hat{y}$ ki taraf point karta hai: diameter ke perpendicular, page ke plane mein, semicircle ke bulge se door. Figure mein orange arrow yeh direction dikhata hai. Answer: $0.72\mathrm{N}$, diameter ke perpendicular.

KYA: wohi shortcut apply karo, lekin ab wire apni start par wapas aati hai. $\vec{F}=I(\oint d\vec{\ell })\times \vec{B}.$ Kisi bhi closed path ke liye, $\oint d\vec{\ell }=\vec{0}$ (tum wahan khatam hote ho jahan shuru hua tha, isliye total displacement zero hai). $\vec{F}=I(\vec{0})\times \vec{B}=\vec{0}.$ Answer: net force zero hai. Har side phir bhi force feel karti hai, lekin opposite contributions cancel ho jaate hain. Jo bachta hai woh hai torque — current-loop torque aur motor ka seed.

KYA: upward magnetic force ko downward weight ke equal set karo, per unit length. KYUN per unit length: dono forces $L$ ke saath scale karte hain, isliye $L$ cancel ho jaata hai aur ek clean condition milti hai. Weight per length: $\frac{mg}{L}=\lambda g$. Magnetic force per length ($\theta =90^{\circ }$ par): $\frac{F}{L}=BI$. Balance: $BI=\lambda g\Rightarrow I=\frac{\lambda g}{B}=\frac{0.050\times 9.8}{0.40}.$ $I=\frac{0.49}{0.40}=1.225\mathrm{A}\approx 1.2\mathrm{A}.$ Answer: $I\approx 1.2\mathrm{A}$, us direction mein flow karte hue jo (Fleming's Left-Hand Rule se, is page ke upar define kiya) force ko upar point karata hai.

KYA: pehle fields add karo (fields superpose hoti hain), phir ek cross product lo. KYUN superpose: magnetic fields vectors ki tarah add hoti hain, isliye wire hamesha sirf total ${\vec{B}}_{\text{tot}}$ "feel" karti hai.

KYA: ek turn ki ek vertical side length $\ell$ ki ek seedhi wire hai jo $\vec{B}$ ke perpendicular hai, isliye use $F_1=BI\ell$ milta hai. Kyunki us side par $N$ turns stack hain, sab same current same taraf carry kar rahe hain, unki forces seedhi upar add ho jaati hain. KYUN $N$ se multiply karein: $N$ wires ek doosre ke paas same field mein same taraf point karte hue lie karti hain, isliye unki equal forces simply sum ho jaati hain — yahi exactly woh reason hai kyun zyada turns galvanometer ko zyada sensitive banate hain. $F=NBI\ell =50\times 0.10\times (2.0\times 10^{-3})\times \mathrm{0.030.}$ Step by step: $50\times 0.10=5.0$; phir $5.0\times 2.0\times 10^{-3}=1.0\times 10^{-2}$; phir $1.0\times 10^{-2}\times 0.030=3.0\times 10^{-4}$. $F=3.0\times 10^{-4}\mathrm{N}.$ Answer: $F=3.0\times 10^{-4}\mathrm{N}$ ek vertical side par. KYUN radial field matter karta hai: kyunki yeh har deflection angle par $\theta =90^{\circ }$ rakhta hai, $F$ (aur isliye torque) $I$ ke proportional rehta hai — yahi woh cheez hai jo galvanometer ka scale linear banati hai.

KYA/KYUN: Loop ki do opposite sides opposite directions mein current carry karti hain, isliye $\vec{F}=I\vec{L}\times \vec{B}$ se unhe equal aur opposite forces milti hain — vector sum ($\oint d\vec{\ell }=0$) zero hai, isliye loop poori tarah kahin push nahi hota. Lekin woh do equal-opposite forces alag lines par act karti hain (woh loop ki width se offset hain), ek couple banati hain. Ek couple torque produce karta hai bina net force ke — yeh loop ko rotate karta hai. Woh rotation, commutator ke saath jo har half-turn par current flip karta hai, electric motor hai.

Perpendicular force, $L=0.40$, $I=5.0$, $B=0.20$? ::: $0.40\mathrm{N}$. $30^{\circ }$ par force, $L=0.60$, $I=3.0$, $B=0.50$? ::: $0.45\mathrm{N}$. $F=0.24$, $I=2.0$, $L=0.30$, $90^{\circ }$ se field? ::: $0.40\mathrm{T}$. Semicircle net force ($R=0.20$, $I=6.0$, $B=0.30$)? ::: $0.72\mathrm{N}$ ($L_{\text{eff}}=2R$ use karo). Floating-wire current ($\lambda =0.050$, $B=0.40$, $g=9.8$)? ::: $\approx 1.2\mathrm{A}$. Galvanometer side force ($N=50$, $B=0.10$, $I=2.0\mathrm{mA}$, $\ell =0.030$)? ::: $3.0\times 10^{-4}\mathrm{N}$. Uniform $B$ mein kisi bhi closed loop par net force? ::: Zero; torque reh sakta hai.